Free Fall

Beginner

Drop a ball and it falls faster and faster. An object is in free fall when gravity is the only force that matters. Near the Earth's surface it gains about 9.81 m/s of downward speed every second, whatever it is made of. This steady gain is called g, the acceleration due to gravity.

Let a stone go from a high bridge. After 1 s it falls at about 10 m/s, after 2 s at about 20 m/s, after 3 s at about 29 m/s: the speed climbs by the same step each second.

Quick check: roughly how fast is it after 5 s? (About 49 m/s.)

The distance grows faster than the speed, because the stone starts from rest. In the first second its average speed is only about 4.9 m/s, so it drops about 4.9 m; after two seconds it has fallen 19.6 m, four times as far. In the animation the faint circles, one every half second, spread further and further apart.

Why do heavy and light things fall together? Gravity pulls a heavier object harder, but a heavier object is also harder to speed up, in exactly the same proportion, so the two effects cancel. In a tube with the air pumped out, a feather and a coin fall side by side. In everyday life air pushes back on a falling object, more and more as it speeds up, so a feather or a parachute soon stops gaining speed.

Gravity does not switch off on the way up. A ball thrown upward loses about 9.81 m/s every second, stops for an instant at the top and falls back.

Quick check: a ball leaves your hand straight up at about 20 m/s. When does it stop rising? Answer: after about 2 s.

Intermediate

Free fall and the choice of sign

Timing a dropped stone is one way to find the depth of a well. An object is in free fall when gravity is the only force acting on it, with air resistance neglected. A ball thrown upward is in free fall too, from the moment it leaves your hand. Near the Earth's surface every freely falling body has the same downward acceleration, $g\approx 9.81m/{\mathrm{s}}^2$, whatever its mass.

The motion is along a vertical line, so choose which way is positive. For a drop, take down as positive and start the clock at the release: the velocity $v$, the acceleration $g$ and the distance fallen $h$ are then all positive. For a throw, take up as positive: the acceleration is then $-g$, and a negative velocity means the object is coming down. Never mix the two in one problem. In the formulas below $h$ is the distance fallen; the animation's $h$ readout shows the height above the ground.

One acceleration for every body

In free fall the velocity changes by the same amount every second, $g=\Delta v/\Delta t$: 9.81 m/s of downward velocity per second, for a coin or a person alike. Gravity pulls with the weight $W=mg$, proportional to the mass, but the mass also resists any change of velocity, and Newton's second law gives $a=W/m=g$. A body ten times heavier is pulled ten times harder and is ten times harder to speed up.

Galileo reached the same conclusion by reasoning. If heavy bodies fell faster, a light stone tied to a heavy one should hold it back, yet the pair is heavier than either and should fall faster still; the contradiction disappears only if all bodies fall alike. In 1971 an Apollo 15 astronaut dropped a hammer and a falcon feather on the airless Moon, where $g\approx 1.62m/{\mathrm{s}}^2$, and they landed together.

The free-fall equations

Since $g$ is constant, the constant-acceleration equations of Motion in a Straight Line apply with $v_0=0$ and $a=g$, down being positive:

• $v=gt$
• $h=\frac{1}{2}g{t}^2\text{(distance fallen from rest)}$
• $v=\sqrt{2gh}$

The speed grows in proportion to the time, the distance with its square: in twice the time an object falls four times as far, because it is also moving faster. The third equation, from $v^2=2gh$, links speed and height without the time. Handy checks: a fall of about 5 m takes 1 s, one of about 20 m takes 2 s.

For a throw, keep the initial velocity $v_0$. With up as positive and $y$ the height above the launch point,
$v=v_0-gt\text{(up positive)}$
$y=v_{0}t-\frac{1}{2}g{t}^2$
A ball thrown up at $v_0$ stops at the top after $v_0/g$, at a height $v_0^2/(2g)$, and is back at the launch point after $2v_0/g$, moving down at the speed $v_0$: the way down mirrors the way up. A throw downward simply has a negative $v_0$.

Worked example: the ball in the animation

How long does the fall take, and how fast does the ball land? The ball is released from rest $h_0=40\mathrm{m}$ above the ground and lands when it has fallen 40 m:
$t=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2\cdot 40}{9.81}}\approx 2.86\mathrm{s}$
$v=gt=9.81\cdot 2.856\approx 28.0m/s$
The third equation agrees, without the time: $\sqrt{2\cdot 9.81\cdot 40}=\sqrt{784.8}\approx 28.0m/s$, about 100 km/h.

These are the readouts at the landing: $t$ stops at 2.9 s (they show one decimal) and $v$ at 28.0 m/s, while the $h$ readout, the height $h_0-\frac{1}{2}g{t}^2$, runs down from 40 m to 0. The blue $v$ arrow grows by the same step every second, and the green $g$ arrow keeps its length throughout.

Halfway through the time, at 1.43 s, the ball has fallen only 10.0 m, a quarter of the drop, and moves at 14.0 m/s; the $h$ readout shows 30.0 m. It passes the halfway height, 20 m, only after 2.02 s, at 19.8 m/s: it crosses the lower half in less than a third of the time. Its average speed, $40/2.856\approx 14.0m/s$, is half the final speed.

Worked example: a ball thrown straight up

You throw a ball straight up at 12 m/s from a hand 1.5 m above the ground, then step aside. How high does it rise, when is it back at hand height, and how fast does it land? Take up as positive and the origin at the hand: $v_0=+12m/s$, acceleration $-9.81m/{\mathrm{s}}^2$.

At the top $v=0$, so $0=12-9.81t$ gives $t\approx 1.22\mathrm{s}$, and the height gained is $v_0^2/(2g)=144/19.62\approx 7.34\mathrm{m}$: the top is 8.84 m above the ground. The ball is back at hand height after $2v_0/g\approx 2.45\mathrm{s}$, moving down at 12 m/s. At $t=2\mathrm{s}$, $v=12-9.81\cdot 2=-7.6m/s$: the minus sign means it is already falling.

On reaching the ground its displacement is $\Delta y=-1.5\mathrm{m}$, and the time is not needed, so use $v^2=v_0^2-2g\Delta y$:
$v^2=12^2-2\cdot 9.81\cdot (-1.5)=144+29.4=173.4{\mathrm{m}}^2/{\mathrm{s}}^2$
It lands at 13.2 m/s downward, $v=-13.2m/s$, after $t=(12+13.2)/9.81\approx 2.57\mathrm{s}$, faster than it was thrown because it ends 1.5 m lower.

Reading the strobe picture and the graphs

The faint circles show the ball every 0.5 s; at the circles labelled 1 s and 2 s it has fallen 4.9 m and 19.6 m. The gaps between successive circles are 1.23, 3.68, 6.13, 8.58 and 11.04 m, in the ratio 1 : 3 : 5 : 7 : 9. This is Galileo's odd-number rule: the distance goes as $t^2$ (1, 4, 9, 16, 25), and successive squares differ by odd numbers. Each gap is $g(\Delta t{)}^2=9.81\cdot 0.5^2\approx 2.45\mathrm{m}$ longer than the one before.

On a velocity–time graph the fall is a straight line through the origin with slope $g$, and the distance fallen is the area under it: for the animation, a triangle 2.86 s wide and 28.0 m/s tall, of area $\frac{1}{2}\cdot 2.856\cdot 28.0\approx 40\mathrm{m}$. The height–time graph, drawn in the expert layer, is a parabola that starts flat and steepens, its slope being the velocity.

Common mistakes

• The acceleration is not zero at the top: only the velocity is, for an instant, while the acceleration stays 9.81 m/s² downward and turns the ball back. Thrown up at 19.62 m/s, a ball stops after 2.0 s and returns after 4.0 s.
• Heavier objects do not fall faster: without air, a 5 kg shot and a 5 g marble both gain 9.81 m/s every second.
• Keep the factor 2 and the square root in $v=\sqrt{2gh}$: a drop of 20 m gives $\sqrt{392.4}\approx 19.8m/s$, not 392 or 196 m/s.
• Do not confuse the distance fallen with the height above the ground: after 1 s the animation's ball has fallen 4.9 m and is 35.1 m up.

Where you meet it

Drop a ruler between a friend's fingers: if it falls 20 cm before they catch it, their reaction time is $\sqrt{2\cdot 0.20/9.81}\approx 0.20\mathrm{s}$. A diver from a 10 m platform reaches the water after about 1.4 s at about 14 m/s, and timing a dropped stone gives the depth of a well. Free fall is also the vertical half of every throw: Projectile Motion adds a steady horizontal motion, Newton's Laws of Motion explains the acceleration, Work, Energy and Power reads $v^2=2gh$ as energy conservation, and Drag and Terminal Velocity adds the air.

Advanced

Inertia, weight and the equivalence principle

Everything in this lesson rests on one experimental fact: in a vacuum, all bodies near the ground fall with the same acceleration, whatever their mass or material. This section shows why that fact is surprising and where it led.

Take up as the positive $y$ direction. Newton's second law for a falling body then reads
$m_i\ddot{y}=-m_{g}g$
Two different masses appear here. On the left, the inertial mass $m_i$ measures how hard the body is to accelerate. On the right, the gravitational mass $m_g$ measures how strongly gravity pulls on it, just as an electric charge measures how strongly an electric field pulls; $g$ is the strength of the Earth's gravity near the ground, about 9.81 m/s². Nothing in Newton's mechanics says that the two masses must be equal.

Experiment says they are. Torsion balances find the ratio $m_g/m_i$ the same for different materials to about one part in $10^{13}$, and in 2022 a satellite comparing test masses of platinum and titanium alloys reached about one part in $10^{15}$. With $m_i=m_g$ the mass cancels from both sides, leaving $\ddot{y}=-g$ for every body: the acceleration of free fall does not depend on what is falling.

Einstein turned this equality into a principle, the equivalence principle. Picture a small closed box falling freely. Every object inside has the same acceleration as the box, so relative to the box nothing accelerates: loose objects float, and no experiment done inside can tell the fall from floating far out in empty space. The box must be small, so that $g$ has the same size and direction throughout it. The equivalence principle is the starting point of general relativity, the theory in which gravity becomes the geometry of spacetime.

This is why astronauts float. At a space station's altitude of about 400 km, $g$ is still about 8.7 m/s², nearly 90 % of its value at the ground. Gravity has not switched off: the crew and the station are falling together, so neither presses on the other.

Integrating the equation of motion

With the mass gone, the whole motion follows from $a_y=-g$. Two integrations give the velocity and the position, and one more step gives the energy.

Take up as positive and start the clock at the release, $t=0$, when the body is at the height $h_0$ with the vertical velocity $v_0$. Because $a_y=d{v}_y/dt=-g$ is constant, one integration gives the velocity; because $v_y=dy/dt$, a second gives the position:
$v_y(t)=v_0+{\int }_0^t(-g)d{t}^{\prime }=v_0-gt$
$y(t)=h_0+{\int }_0^t{v}_y(t^{\prime })d{t}^{\prime }=h_0+v_{0}t-\frac{1}{2}g{t}^2$
Each integration brings one constant, fixed by the starting state: $v_0$ in the first, $h_0$ in the second. Gravity itself supplies only the term $-\frac{1}{2}g{t}^2$, which bends the graph of $y(t)$ into a parabola, and it bends every fall in the same way.

In the animation the ball is dropped from rest, $v_0=0$, at $h_0=40\mathrm{m}$. It reaches the ground, $y=0$, when $\frac{1}{2}g{T}^2=h_0$, that is at
$T=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2\cdot 40}{9.81}}\approx 2.86\mathrm{s}$
Its velocity is then $v_y=-gT\approx -28.0m/s$, the minus sign meaning downwards.

Often the time does not matter, only how the speed depends on the height. The chain rule removes $t$: $a_y=d{v}_y/dt=(d{v}_y/dy)(dy/dt)=v_{y}d{v}_y/dy$. So $v_{y}d{v}_y=-gdy$, and integrating from the release gives
$v_y^2=v_0^2-2g(y-h_0)$
Multiply by $m/2$ and rearrange: $\frac{1}{2}m{v}_y^2+mgy=\frac{1}{2}m{v}_0^2+mg{h}_0$. The left side, kinetic plus potential energy, keeps its starting value: energy is conserved. Each kilogram of the animation's ball starts with $g{h}_0=9.81\cdot 40=392.4J/kg$ of potential energy and lands with $\frac{1}{2}{v}_y^2=\frac{1}{2}\cdot 784.8=392.4J/kg$ of kinetic energy.

Reading the velocity and height graphs

The two formulas are easiest to understand as graphs. Reading slopes and areas on them is the quickest check of any motion with constant acceleration.

Legend

$t$	Time since the release.	$\mathrm{s}$
$T$	Time of the impact, where the fall ends.	$\mathrm{s}$
$v_y$	Vertical velocity, up positive: the height of the upper line, and the slope of the tangent to $h(t)$.	$m/s$
$gT$	Speed gained in the whole fall; the upper line ends at $v_y=-gT$.	$m/s$
$\Delta t$	Any time interval: the horizontal leg of the slope triangle.	$\mathrm{s}$
$\Delta v_y$	Change of $v_y$ over $\Delta t$: the vertical leg. The slope is $\Delta v_y/\Delta t=-g$.	$m/s$
$\Delta h$	Change of height in the whole fall: the signed area between the $v_y$ line and the $t$ axis.	$\mathrm{m}$
$h$	Height above the ground.	$\mathrm{m}$
$h_0$	Release height, at $t=0$.	$\mathrm{m}$

Figure 1 shows the animation's drop as two graphs, one above the other, on the same time axis $t$. The vertical dashed line at the impact time $T\approx 2.86\mathrm{s}$ joins them.

Look first at the upper graph, $v_y(t)=-gt$: a straight line from zero down to $-gT\approx -28.0m/s$. Its slope is the acceleration. The small triangle shows that over any interval $\Delta t$ the velocity changes by $\Delta v_y=-g\Delta t$, and the slope of the whole line, $-28.01/2.856\approx -9.81m/{\mathrm{s}}^2$, is $a_y=-g$. Because the slope is the same everywhere, so is the acceleration: on a graph, constant acceleration means a straight velocity line.

Now look at the shaded triangle between the line and the time axis. Its area is the change of height $\Delta h$, because each thin vertical strip of it has the width $dt$ and the height $v_y$, and $v_{y}dt$ is the distance moved in that time. The triangle lies below the axis, so its area counts as negative: $\Delta h=\frac{1}{2}\cdot 2.856\cdot (-28.0)\approx -40\mathrm{m}$, the whole drop.

The lower graph is the height $h(t)$, which during the fall is $y(t)=h_0-\frac{1}{2}g{t}^2$: a parabola opening downwards. Its slope at each instant is the current $v_y$, read off the upper graph. At the release $v_y=0$, so the curve leaves the level $h_0=40\mathrm{m}$ horizontally. At the impact $v_y=-28.0m/s$, so the curve is steepest there; the tangent drawn at $T$, labelled $v_y$, has the slope $-gT$, the value at which the upper line ends. The rate at which the slope changes, the bending of the curve, is the second derivative, $-g$ all the way down.

The parabola also shows how unevenly the drop spends its time. At half time, $T/2=1.43\mathrm{s}$, the ball has fallen only $\frac{1}{2}\cdot 9.81\cdot 1.428^2=10.0\mathrm{m}$ and is still at 30 m, three quarters of the way up; the remaining 30 m take the second half. At $T$ both graphs stop: the ground pushes back, and the free fall is over.

Where g comes from

So far $g$ has been a given number. Newton's law of universal gravitation explains its value, and shows why it is only approximately constant.

A spherical Earth of mass $M$ and radius $R$ attracts a body of mass $m$ at its surface with the force $GMm/R^2$, as if all its mass sat at the centre. Dividing by $m$ gives the acceleration
$g=\frac{GM}{R^2}$
With $G=6.674\times 10^{-11}\mathrm{N}{\mathrm{m}}^2/k{\mathrm{g}}^2$, $M=5.97\times 10^{24}\mathrm{kg}$ and $R=6371\mathrm{km}$ it comes to 9.82 m/s². The measured value differs slightly from place to place, because the Earth spins, is flattened at the poles and is not uniform inside.

At a height $h$ above the surface the distance from the centre is $R+h$, so $g$ becomes $GM/(R+h{)}^2\approx (GM/R^2)(1-2h/R)$ when $h\ll R$: smaller by a fraction of about $2h/R$. For the animation's 40 m that fraction is $80/(6.371\times 10^6)\approx 1.3\times 10^{-5}$, about 13 parts per million. This is why $g$ looks constant: everyday heights are tiny compared with $R$.

An orbit is a fall that keeps missing the ground

The same $g$ that brings a dropped ball down also keeps a satellite up. Figure 2 shows how, by comparing two small distances.

Legend

$O$	Centre of the Earth.
$R$	Radius of the Earth.	$\mathrm{m}$
$\vec{v}$	Velocity at the start, horizontal, along the tangent.	$m/s$
$vt$	Distance along the straight line in the time $t$, without gravity.	$\mathrm{m}$
$\vec{g}$	Acceleration of gravity, pointing at the centre $O$.	$m/{\mathrm{s}}^2$
$\frac{1}{2}g{t}^2$	Distance gravity pulls the ball below the straight line in the time $t$.	$\mathrm{m}$

In Figure 2 a ball is thrown sideways from just above the ground with the velocity $\vec{v}$. Without gravity it would move along the dashed straight line, a tangent to the Earth, and after a short time $t$ it would be at the pale ball, a distance $vt$ from its start. Gravity adds a displacement of about $\frac{1}{2}g{t}^2$ towards the centre $O$, the gold segment. So after the time $t$ the ball is not at the pale ball but at the solid ball just below it, on the curved path; there the arrow $\vec{g}$ points at $O$.

The ground also falls away from the dashed line, because the Earth is round. The tangent is at right angles to the radius $R$, so by Pythagoras a point a distance $x$ along it is $\sqrt{R^2+x^2}$ from $O$, and its height above the surface is
$\sqrt{R^2+x^2}-R\approx \frac{x^2}{2R}$
for $x\ll R$. With $x=vt$, the ground lies $(vt{)}^2/(2R)$ below the line.

Now compare the two drops. If the ball falls less than the ground, it climbs away from the ground; if more, it lands; if exactly as much, it stays at the same height above the ground. Since the same holds at every point of its path, it then circles forever. The condition is
$\frac{1}{2}g{t}^2=\frac{(vt{)}^2}{2R}$
The time cancels, leaving $v^2/R=g$: gravity supplies exactly the centripetal acceleration $v^2/R$ that a circle of radius $R$ needs. Near the surface this takes $v=\sqrt{gR}=\sqrt{9.81\cdot 6.371\times 10^6}\approx 7.9km/s$. An orbit is free fall with enough sideways speed for the ground to keep curving away underneath.

The same reasoning passed Newton's decisive test, the Moon. If gravity weakens as $1/r^2$ with the distance $r$ from the centre, then at the Moon's distance of about 60 Earth radii it should give an acceleration of $g/60^2\approx 2.7\times 10^{-3}m/{\mathrm{s}}^2$. The Moon actually circles at 384,000 km once every 27.3 days, which requires a centripetal acceleration of $2.72\times 10^{-3}m/{\mathrm{s}}^2$: the two agree.

The same agreement can be pictured as a distance fallen. In one second the Moon falls $\frac{1}{2}\cdot 2.72\times 10^{-3}\cdot 1^2\approx 1.4\times 10^{-3}\mathrm{m}$, about 1.4 mm, below the straight line it would otherwise follow. That is the $\frac{1}{2}g{t}^2$ of Figure 2 with the Moon's much smaller $g$; an apple near the ground falls 4.9 m in the same second.

Falling through air

In air, the rule that all bodies fall alike breaks down, and it breaks down in a predictable way. Figure 3 shows the effect before the equations explain it.

Legend

$t$	Time since the release.	$\mathrm{s}$
$v$	Speed of the ball, measured downwards.	$m/s$
$gt$	Speed after a time $t$ with no air: the dashed straight line.	$m/s$
$v_t$	Terminal speed of the lighter ball. The heavier ball's is $2v_t$, since $v_t\propto \sqrt{m}$.	$m/s$
$m$	Mass of the lighter ball; the heavier one has $4m$.	$\mathrm{kg}$

Figure 3 plots the speed $v$ against the time $t$ for two balls of the same size and shape, one of mass $m$ and one of mass $4m$. The dashed straight line is $v=gt$, the fall without air. Both curves leave the origin along it, because at low speed the drag is negligible. Then they bend away and level off at their own terminal speeds, marked by the horizontal dashed lines: $v_t$ for the lighter ball and $2v_t$ for the heavier one. The heavier ball follows the straight line for longer and ends up twice as fast. The rest of this section derives the curves and the factor 2.

Drag is the force the air exerts on a body moving through it, directed against the motion and growing with the speed. For everyday sizes and speeds it is close to
$F_d=\frac{1}{2}\rho C_{d}A{v}^2$
where $\rho$ is the density of the air, $A$ the cross-sectional area facing the flow, and $C_d$ the drag coefficient, a pure number set by the shape, about 0.5 for a sphere. For dust and fog droplets, which are tiny and slow, the drag is proportional to $v$ instead.

Take down as positive. Newton's second law, with the weight pulling down and the drag pushing up, reads
$m\frac{dv}{dt}=mg-\frac{1}{2}\rho C_{d}A{v}^2$
At the start $v=0$, the drag is zero and the acceleration is $g$. As the speed grows the drag catches up with the weight and the acceleration fades. The speed approaches the value at which the two forces balance and $dv/dt=0$, the terminal velocity:
$v_t=\sqrt{\frac{2mg}{\rho C_{d}A}}$

Written with $v_t$, the equation of motion becomes $dv/dt=g(1-v^2/v_t^2)$. Its solution starting from rest is
$v(t)=v_t\tanh \left(\frac{gt}{v_t}\right)$
You can check it by differentiating: since $d(\tanh x)/dx=1-{\tanh }^{2}x$, the derivative is $g(1-v^2/v_t^2)$. For small $t$, $\tanh x\approx x$ and $v\approx gt$, the no-air line; for $t\gg v_t/g$, $\tanh \to 1$ and $v\to v_t$. So $v_t/g$ is the time scale on which the curve levels off.

To get the speed after a fall through a distance $s$, use the chain rule again: $dv/dt=vdv/ds=\frac{1}{2}d(v^2)/ds$. The equation for $v^2$ is then
$\frac{d(v^2)}{ds}=2g\left(1-\frac{v^2}{v_t^2}\right)$
so $v^2$ approaches $v_t^2$ exponentially, over a distance scale $v_t^2/(2g)$:
$v^2=v_t^2\left(1-e^{-2gs/v_t^2}\right)$
For a short fall, $s\ll v_t^2/(2g)$, the exponential is close to $1-2gs/v_t^2$, and the result reduces to the no-air $v^2=2gs$.

Now the factor 2. At fixed size and shape, $\rho$, $C_d$ and $A$ are the same for both balls, so $v_t\propto \sqrt{m}$: four times the mass gives twice the terminal speed. It also doubles the time scale $v_t/g$, so the heavier ball stays close to $v=gt$ for twice as long. That is the second curve of Figure 3.

Drop a tennis ball, the lighter ball of Figure 3, from the animation's 40 m. Take $m=57\mathrm{g}$, a diameter of 6.7 cm (so $A=\pi \cdot 0.0335^2\approx 3.53\times 10^{-3}{\mathrm{m}}^2$), $C_d\approx 0.55$ and $\rho =1.2kg/{\mathrm{m}}^3$. Its terminal velocity is
$v_t=\sqrt{\frac{2\cdot 0.057\cdot 9.81}{1.2\cdot 0.55\cdot 3.53\times 10^{-3}}}\approx 22m/s$
and it lands at 19.7 m/s after 3.26 s, instead of 28.0 m/s after 2.86 s without air. A ball of the same size four times as heavy has twice the terminal velocity, about 44 m/s, and lands at 25.4 m/s after 2.95 s. So air does separate heavy from light, through $v_t\propto \sqrt{m}$ at fixed size and shape.

For comparison, a skydiver falling belly-down levels off near 55 m/s (about 200 km/h), and a raindrop 1 mm across at about 4 m/s. At terminal velocity the acceleration is zero, not $g$: the body falls at constant speed, with the drag balancing the weight.

Worked problem: meeting the falling ball

At the instant the ball of the animation is released from 40 m, a second ball is thrown straight up from the ground directly below it at 28.0 m/s, the speed at which the first one lands. When and where do they meet?

Legend

$t$	Time since the release of the first ball.	$\mathrm{s}$
$T$	Time at which the first ball lands and the second reaches its highest point.	$\mathrm{s}$
$\frac{T}{2}$	Time at which the balls meet.	$\mathrm{s}$
$y$	Height above the ground, up positive.	$\mathrm{m}$
$h_0$	Release height of the first ball, and the top of the second ball's flight.	$\mathrm{m}$
$\frac{3}{4}{h}_0$	Height at which the balls meet.	$\mathrm{m}$
$y_1$	Height of the dropped ball.	$\mathrm{m}$
$y_2$	Height of the ball thrown up from the ground.	$\mathrm{m}$

Figure 4 plots both heights against time, with up as positive and the origin on the ground. The first ball is dropped from $h_0$; the second is thrown up with $v_0=28.0m/s$:
$y_1=h_0-\frac{1}{2}g{t}^2,y_2=v_{0}t-\frac{1}{2}g{t}^2$
Follow $y_1$: it leaves $h_0$ horizontally and reaches the ground at $T$. Follow $y_2$: it starts on the ground, and since $v_0=gT$, its velocity $v_0-gt$ falls to zero exactly at $T$. So the second ball reaches the top of its flight at $T$, at the height $v_{0}T-\frac{1}{2}g{T}^2=\frac{1}{2}g{T}^2=h_0$, where the first ball started.

The two curves are mirror images of each other about the dashed line at $T/2$: $y_2(t)=y_1(T-t)$. To check, use $h_0=\frac{1}{2}g{T}^2$: then $y_1(T-t)=h_0-\frac{1}{2}g(T-t{)}^2=gTt-\frac{1}{2}g{t}^2$, which is $y_2(t)$. The balls meet where the curves cross, and by the symmetry that crossing lies on the dashed line.

The algebra is even shorter. The $\frac{1}{2}g{t}^2$ terms are identical, so they cancel from the gap between the balls: $y_1-y_2=h_0-v_{0}t$. The gap closes at a constant 28.0 m/s, as if there were no gravity, and vanishes at $t=h_0/v_0=40/28.01\approx 1.43\mathrm{s}$, exactly $T/2$.

At that moment both balls are at $y=40-\frac{1}{2}\cdot 9.81\cdot 1.428^2=30.0\mathrm{m}$, the level $\frac{3}{4}{h}_0$ marked in Figure 4. The first ball moves down at $gT/2=14.0m/s$ and the second up at $28.0-14.0=14.0m/s$. The equal speeds are no coincidence: the second ball's flight is the first one's played backwards, which is what the mirror symmetry of Figure 4 says.

A slower throw meets later and lower. The meeting time $h_0/v_0$ cannot exceed the 2.86 s the first ball takes to land, so the throw needs at least $v_0=h_0/T=14.0m/s$, the average speed of the fall; at exactly that speed both balls reach the ground together.

Seen from the first ball, gravity has disappeared: both balls have the same acceleration $-g$, so in the first ball's freely falling frame the second ball approaches in a straight line at the constant speed $v_0$. This is the equivalence principle of the first section at work. The same reasoning explains why, in Projectile Motion, a dart aimed straight at a target that drops at the moment of firing hits it whatever its launch speed, provided it arrives before the target lands.

Limits of validity

The ideal model of free fall makes three assumptions: $g$ is uniform, there is no air, and the Earth does not rotate. The numbers above show how far each one holds.

A uniform $g$ is excellent over a 40 m drop, where $g$ changes by 13 parts per million. A fall from orbital heights, however, needs $g=GM/r^2$ at each distance $r$ from the centre. Neglecting the air is fine only while $v\ll v_t$: from 40 m, the heavier ball of the previous section lands 9 % short of the ideal 28.0 m/s, and the tennis ball 30 % short.

The Earth's rotation makes the ground a slightly non-inertial frame, one that is itself accelerating. Part of gravity goes into keeping the ground on its daily circle, which reduces the effective $g$ by up to 0.034 m/s² at the equator. Together with the Earth's flattening, this makes $g$ range from 9.78 m/s² at the equator to 9.83 m/s² at the poles.

Rotation also deflects a fall eastward: the top of a tower moves east slightly faster than its foot, and a ball dropped from it keeps that extra speed. The animation's ball, dropped at a latitude of 45°, would land about 4 mm east of the point directly below its release.

Relativity adds only tiny corrections at these speeds: at 28 m/s, $v^2/c^2\approx 9\times 10^{-15}$. Gravity also affects time itself: a clock 40 m up runs faster than one on the ground by the fraction $gh/c^2\approx 4.4\times 10^{-15}$. This shift was first measured around 1960 with gamma rays sent up a 22.5 m tower, and optical clocks now detect it over height differences of less than a metre.

Where this leads

Free fall connects to most of mechanics. Projectile Motion combines it with a uniform horizontal motion; Newton's Laws of Motion explains the acceleration as weight divided by mass; Work, Energy and Power turns $v^2=2gh$, the speed after a drop of height $h$ from rest, into $mgh=\frac{1}{2}m{v}^2$. Gravitation and Orbits extends $g=GM/r^2$ to planets and satellites, Drag and Terminal Velocity treats the air properly, and Simple Harmonic Motion shows how the period of a pendulum depends on $g$, which is how $g$ was measured for centuries. Gravity as Curved Spacetime builds a whole theory on the fact that all bodies fall alike.